Question: Is ${54091}$ divisible by $3$ ?
Explanation: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {54091}= &&{5}\cdot10000+ \\&&{4}\cdot1000+ \\&&{0}\cdot100+ \\&&{9}\cdot10+ \\&&{1}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {54091}= &&{5}(9999+1)+ \\&&{4}(999+1)+ \\&&{0}(99+1)+ \\&&{9}(9+1)+ \\&&{1} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {54091}= &&\gray{5\cdot9999}+ \\&&\gray{4\cdot999}+ \\&&\gray{0\cdot99}+ \\&&\gray{9\cdot9}+ \\&& {5}+{4}+{0}+{9}+{1} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first four terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${54091}$ is divisible by $3$ if ${ 5}+{4}+{0}+{9}+{1}$ is divisible by $3$ Add the digits of ${54091}$ $ {5}+{4}+{0}+{9}+{1} = {19} $ If ${19}$ is divisible by $3$ , then ${54091}$ must also be divisible by $3$ ${19}$ is not divisible by $3$, therefore ${54091}$ must not be divisible by $3$.